Amit & Deepali Sinha Foundation

Colliding Particles Share One Line

Irodov 1.5DifficultyUniqueness

Problem Irodov 1.5

Two particles, labeled 11 and 22, move in a plane, each with its own constant velocity vector, v1\vect v_1 and v2\vect v_2. At some initial instant their position vectors, measured from a common fixed origin, are r1\vect r_1 and r2\vect r_2. Find the relation these four vectors must satisfy for the two particles to arrive at the same point at the same time (that is, to collide).

The four given vectors at the initial instant: r_1 and r_2 locate the two particles from the common origin O, and each carries its own constant velocity, drawn in a generic direction.

The four given vectors at the initial instant: r1\vect r_1 and r2\vect r_2 locate the two particles from the common origin OO, and each carries its own constant velocity, drawn in a generic direction.

Given
  • r1\vect r_1, r2\vect r_2
  • v1\vect v_1, v2\vect v_2
Find
the relation among r1,r2,v1,v2\vect r_1,\vect r_2,\vect v_1,\vect v_2 for a collision

Solution

Step 1Write each particle’s position as a function of time

With constant velocity, each particle’s position vector is its initial radius vector plus velocity times elapsed time.

r1(t)=r1+v1tr2(t)=r2+v2t(1) \begin{aligned} \vect r_1(t) &= \vect r_1+\vect v_1 t\\ \vect r_2(t) &= \vect r_2+\vect v_2 t \end{aligned} \tag{1}
Step 2Impose the collision condition

Setting r1(t∗)=r2(t∗)\vect r_1(t^\ast)=\vect r_2(t^\ast) for some common time t∗t^\ast equates the two paths (1); collecting the position terms on one side then isolates the separation.

r1+v1t∗=r2+v2t∗\vect r_1+\vect v_1 t^\ast = \vect r_2+\vect v_2 t^\ast
r1−r2=(v2−v1)t∗\vect r_1-\vect r_2 = (\vect v_2-\vect v_1)t^\ast
(2)
The configuration describes. Each gold arrow is the displacement its particle travels over the common time t^, drawn at full length, so both arrowheads arrive together at the shared point C.

The configuration (2) describes. Each gold arrow is the displacement its particle travels over the common time t∗t^\ast, drawn at full length, so both arrowheads arrive together at the shared point CC.

Step 3Extract the direction relation

Equation (2) says r1−r2\vect r_1-\vect r_2 is v2−v1\vect v_2-\vect v_1 scaled by the positive number t∗=∣r1−r2∣/∣v2−v1∣t^\ast=|\vect r_1-\vect r_2|/|\vect v_2-\vect v_1|, so the two vectors share a direction. Dividing each side by its own magnitude isolates that shared unit vector.

r1−r2∣r1−r2∣=(v2−v1)t∗∣v2−v1∣t∗\frac{\vect r_1-\vect r_2}{|\vect r_1-\vect r_2|} = \frac{(\vect v_2-\vect v_1)t^\ast}{|\vect v_2-\vect v_1|t^\ast}
r1−r2∣r1−r2∣=v2−v1∣v2−v1∣\dfrac{\vect r_1-\vect r_2}{|\vect r_1-\vect r_2|} = \dfrac{\vect v_2-\vect v_1}{|\vect v_2-\vect v_1|}
(3)
Check

Setting v1=v2\vect v_1=\vect v_2 makes the right side undefined, matching the physical fact that two particles moving with identical velocity never close their initial separation at all, so a finite t∗t^\ast cannot exist unless r1=r2\vect r_1=\vect r_2 already. Both sides are honest unit vectors regardless, so the relation constrains direction only, never magnitude or the collision time itself.

Remark

The relation says the collision condition depends only on the relative velocity v2−v1\vect v_2-\vect v_1, not on v1\vect v_1 and v2\vect v_2 separately: in the frame of particle 1, particle 2 simply travels in a straight line at v2−v1\vect v_2-\vect v_1 starting from r2−r1\vect r_2-\vect r_1, and a collision happens exactly when that straight line is aimed directly back at the origin of that frame, particle 1 itself. This is the same ”aim at the target” idea behind pursuit and interception problems.